Question

Find the equations of tangents to the curve y=x${}^4$ which are drawn from the point $(2,0)$

• A. Only $y-{\left(\frac{8}{3}\right)}^4=4{\left(\frac{8}{3}\right)}^3(x-\frac{8}{3})$
• B. Only $y=0$
• C. Both A and B
• D. Only $y-\left(\frac{8}{3}\right)=4\left(\frac{8}{3}\right)(x-(\frac{8}{3}{)}^4)$

Answer 1

The correct option is C Both A and B

Given, $y=x^4$

$\therefore \frac{dy}{dx}=4x^3$.
Now equation of the tangent through $(2,0)$.
$\frac{y}{x-2}=4x^3$
$y=4x^3(x-2)$
$y=4x^4-8x^3$
$x^4=4x^4-8x^3$
$3x^4-8x^3=0$
$x^3[3x-8]=0$
$x=0$ and $x=\frac{8}{3}$.
If $x=0$ then $y=0$.
$\frac{y-0}{x-2}=\frac{y-0}{x-0}$
$xy=xy-2y$
$-2y=0$
$\therefore y=0$ ... (x-axis).
Now considering $x=\frac{8}{3}$ we get
$y=\frac{4096}{81}$
Hence
$\frac{y-0}{x-2}=\frac{y-\frac{4096}{81}}{x-\frac{8}{3}}$
$xy-\frac{8y}{3}=xy-\frac{4096x}{81}-2y+\frac{8192}{81}$
$-\frac{8y}{3}+2y=-\frac{4096x}{81}+\frac{8192}{81}$
$\frac{2y}{3}=-\frac{4096x}{81}+\frac{8192}{81}$